Answer
$0.0548$
Work Step by Step
$\text{z-score}=\frac{\text{data item-mean}}{\text{standard deviation}}$
Hence here the z-score is: $\frac{84-95}{7}\approx-1.6$
Then, using the table: $P(x\leq 89)\approx P(z\leq-1.6)=0.0548$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 11 Data Analysis and Statistics - Chapter Review - Page 785: 13
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